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Predicting The Proton!

By!

Ian Beardsley!

of 2 70

Section 1.0: The Radius of a Proton!

Proton Inertia…………………………………………..5!

The Constant k…………………………………………9!

Determining the Proton Radius………………………13!

Discussion………………………………………………17!

Section 2.0 The Charge of a Proton!

Proton Charge…………………………………………..19!

Ionization Energy………………………………………..22!

Section 3.0 The Planets and Atoms!

Planets Mirror Atoms…………………………………..26!

Section 4.0 Proton Seconds!

Rigorous Formulation of Proton-Seconds…………..28!

Theory of Compounds…………………………………31!

The Computation……………………………………….32!

The Dynamic Function………………………………….32!

Section 6.0 The Planets as Six-Fold Symmetry!

Orbital Velocity…………………………………………..38!

Eigenvalues and Eigenvectors…………………………51!

Conclusion……………………………………………….58!

Appendix 1………………………………………………60!

Appendix 2………………………………………………61!

Appendix 3……………………………………………….65!

Appendix 4……………………………………………….68"

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Abstract!

We predict the existence of the proton, its radius and charge as a consequence of space which

we show is responsible for inertia, and ﬁnd everything follows from six-fold symmetry, including

the hydrocarbons the skeletons of life. To do this we introduce the idea intermediary mass and

shows that one second is a natural unit, not just on the order of the atoms but on the order of

the solar system. It follows that the solar system is probably mirroring the atom in that we see it

can be described as an equation of state for the Earth-Moon-Sun system that equates to an

equation of state for the proton. We introduce the idea of proton-seconds which suggests there

might be some use in considering chemical equation as algebraic equations as well. We then

present the six-fold nature of the solar system further demonstrating it connection to the atom."

of 4 70

Section 1.0 The Radius of a Proton

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Proton Inertia

The natural constants measure the properties of space and time. We can list some of them here:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

Gravity is a property of space measured by the universal constant of gravity, G:

Equation 1.1

Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons

and neutrons) we will suggest given by:

Equation 1.2.

Which describes mass per meter over time, which is:

Equation 1.3:

It must be adjusted by the fine structure constant . It is my guess the factor should be which

is 18,769.:

Equation 1.4.

Because the fine structure constant squared is the ratio of the potential energy of an electron in

the first circular orbit to the energy given by the mass of an electron in the Bohr model times the

speed of light squared, that is it represents the ground state. It is

Equation 1.5.

Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a

hypersphere. Thus we consider the surface area of a proton, :

Equation 1.6

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G:6.67408 × 10

−11

c : 299,792, 459m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

G = 6.674E − 11N

= 6.674E − 11

⋅ kg

kg ⋅ s

1.82E − 16

kg ⋅ s

(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s/m)

= 4π r

= 8.72E − 30m

of 6 70

We take the square root to get meters:

Equation 1.7

We multiply that with the value we have in equation 1.4:

Equation 1.8

We find that the mass of a proton realizes if we divide this by six seconds:

Equation 1.9

That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:

Equation 1.10

Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the

skeletons of organic matter, the so-called hydrocarbons:

Equation 1.11

Equation 112

If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of

the elements in the periodic table occur for dividing by something less that 1-second. It seems

the duration of a second is natural. If it is, since it was formed by a calendar based on

reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon

orbital mechanics. I find it is, that (See appendix 1 and appendix 2):

That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic

energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by

very small amounts over millions of years. The solar system has evolved towards this since the

explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive

the second in terms of something else. For now we have the mass of a proton as:

1.13

I then developed the concept of proton-seconds where time is associated with atoms through the

number of protons they have.

= 2.953E − 15m

(2.953E − 15m)(3.416E − 12

kg ⋅ s

) = 1.009E − 26kg ⋅ s

1E − 26

6secon d s

= 1.67E − 27kg

1E − 26kg ⋅ s

1secon d

= 1E − 26kg = 6proton s = 6m

1E − 26kg ⋅ s

6secon ds

1E − 26kg ⋅ s

1secon d

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

18α

4πh

of 7 70

This way of looking at things is to say matter is that which has inertia. This means it resists

change in position with a force applied to it. The more of it, the more it resists a force. We

understand this from experience, but what is matter that it has inertia? In this analogy we are

suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As

such there has to be a normal vector holding the higher dimensional sphere in a lower

dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in

the plane there should be a force countering it proportional to the normal holding it in a lower

dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a

sphere. This counter force would be experienced as inertia. Since Planck’s constant h is a

measure of energy over time where space and time are concerned it must play a role. Of course

the radius of a proton plays a role since squared and multiplied by it is the surface area of our

proton embedded in space. The gravitational constant is force produced per kilogram over a

distance, thus it is a measure of how the surrounding space has an effect on the proton giving it

inertia. The speed of light c has to play a role because it is the velocity at which events are

separated through time. The mass of a proton has to play a role because it is a measurement of

inertia itself. And alas the fine structure constant describes the degree to which these factors

have an effect. We see the inertia then in equation 6 is six protons over 1 second, by dimensional

analysis.

Equation 1.14

We find six seconds gives 1 proton is hydrogen:

Equation 1.15

4π

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

Fig. 1

of 8 70

For time t greater than 6 seconds we have fractional protons. For t<6 we the have other

elements.

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave pure number. We

make a program that looks for close to whole number solutions so we can create a table of values

for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. Here is the code for the program:

h 4π r

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#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?):

");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We see the radius of a proton is given by carbon (1 second):

Equation 1.16

The experimental radius of a proton is:

Equation 1.17

The Constant k

But we need to derive the second because our result is evaluated at one second. Warren

Giordano writes in his paper The Fine Structure Constant And The Gravitational Constant: Keys

To The Substance Of The Fabric Of Space, March 21, 2019:

In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic

equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where

‘ ’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational

constant numerically, but neglecting any units.

Let’s do that

4πh

= 8.288587 × 10

−16

m = 0.829f m

= 0.833

0.014f m

1 + α

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(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js

And it works, G is:

G=6.67408E-11 N(m2/kg2)

Let us reformulate this as by introducing Avogadro’s number because multiplying Planck’s

constant ‘ ’ by ‘ ’ is G if multiplied by and Avogadro’s number is

Equation 1.18

We just need the right units. We can get these:

Since grams and atom cancel we can work in grams even though our equations are in kilograms.

Let us not write H, since formally it is grams per mole of hydrogen but write

We have:

Or,…

Equation 1.19

Where

Equation 1.10

Let us say we were to consider Any Element say carbon . Then in general

Equation 1.11

We have

and

Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a

molar mass of 12. We have

1 + α

= Avaga dr o′ s − Nu m ber = 6.02E 23atom s /gram

(1 + α)

H = 6.0003

⋅ s

H = 6.02E 23

atom s

gr a m

⋅

1gr a m

atom

= 6.02E 23

ℍ = 1

gr a m

atom

(1 + α)

⋅ N

ℍ = 6.0003kg

⋅

h(1 + α)N

ℍ = 6Gx

x = 1.00kg

𝔼

ℂ

(1 + α)

⋅ N

𝔼 = 6.0003kg

⋅

ℂ =

6gr a m s

6proton s

6(6E 23proton s)

6gr a m s

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12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus

And it follows that

We see in general since the atomic number Z is the number of protons in an atom that in general

this holds for all elements because

And,

Therefore we always have:

Equation 1.12

is not molar mass, and that is a variable determined by ; it is a mole of atoms multiplied

by the number of protons in . Thus we have created from the mole of atoms an equation of

state for the periodic table. We know that

Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:

Eq. 1.13

This brings up an interesting question: while we have masses characteristic of the microcosmos

like protons, and masses characteristic of the macrocosmos, like the upper limit for a star to

become a white dwarf after she novas (The Chandrasekhar limit) which is 1.44. More mass than

that and she will collapse. We do not have a characteristic mass of the intermediary world where

we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find

that mass let us take the geometric mean between the mass of a proton and the mass of 1.44

solar masses. We could take the average, or the harmonic mean, but the geometric mean is the

squaring of the proportions, it is the side of a square with the area equal to the area of the

rectangle with these proportions as its sides. We have:

ℂ = 6E 23

(1 + α)

⋅ N

ℂ = 6.0003kg

⋅

𝔼

Z ⋅ 6E 23proton s

Z ⋅ gram s

𝔼 =

Z ⋅ gram s

Z ⋅ pr oton s

⋅ 𝔼 = 6E 23

𝔼

(1 + α)

⋅ N

𝔼 ⋅ v

= 422.787kg

⊙

= 1.98847E 30kg

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We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .

We have the intermediary mass is:

Eq. 1.14

All we really need to do now is divide equation 1.13 by equation 1.14 and we get an even number

that is the six of our six-fold symmetry.

Eq. 1.15

The six of our six-fold symmetry.

We have something very interesting here. We have

This is:

Equation 1.16

Where k is a constant, given

Equation 1.17

We can explicitly write the constant k:

Equation 1.18

It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf

will not have its gravity overcome the electron degeneracy pressure and collapse. The non-

relativistic equation is:

Equation 1.19

Let us approximate 0.77 with 3/4. Since we have our constant

And

Then

= 1.67262E − 27kg

= (2.8634E 30)(1.67262E − 27) = 69.205kg

(1 + α)

⋅ N

𝔼 ⋅ v

= 6.1092 ≈ 6

69.205kg

6kg

= 6

k v

= 6

k =

800

k =

(1 + α)

⋅ N

𝔼

M ≤ 0.77

ℏ

= 1.41 ⊙

k =

1 + α

⋅ N

𝔼

= Mm

of 13 70

Equation 1.20

Since our constant k in terms the Chandrasekhar limit is

Equation 1.21

Determining the Proton Radius

We consider two paths, one with velocity c in one medium, the other with velocity v in another.

In order to go from one point two another over two paths, the refraction is such that the sine of

the angle of incidence equals the sine of the angle of refraction.

We have the two paths are travelled in a time t:

This is the mysterious Nature of reality, the path of least time is taken. This falls under the

general heading The Principle of Least Action, attributed to firstly the French Natural

Philosopher and mathematician Louis Maupertuis of the early eighteenth century. I say

mysterious, because as it is said in physics, for something to know the path of least action and

take it, it is as if it has explored first all paths between A and B to know which one would be the

path of least action. Everything in physics comes to this principle. It is called a principle, not a

theory, law, or rule. Yet it seems to be the way Nature behaves, and it is mysterious. Richard

Feynman applied it to quantum mechanics, probably because the mysterious Planck’s constant

that governs quantum mechanics is in Joule-seconds, energy over time, and this is the terms in

which action is formulated mathematically.

In our scenario here we regard matter, the proton in particular, as the cross-section of a

hypersphere. Our two mediums are hyperspace and space and the least action principle applies

in the same mathematical form. This is abstract cosmology, that really says the underlying

mathematics is common to all systems, that in effect they are manifestations of one another. In

this case the principle of least action.

We have

We make the approximation

ℏ

1/2

ℏ = h /2π

k =

8π

(1 + α) ⋅ N

𝔼

sin α

sin β

t = r

c + v

c v

c v − v = c

c v = c

of 14 70

So that

We are saying t=6 seconds is the proton, and r is the radius of a proton r=0.833E-15m. Thus

The radius of a hydrogen atom is .

And we have our one second as a natural constant with respect to the atom. We see it occurs at

t/2 which is at half the radius of a hydrogen atom, which is because we would think we are

looking at one hydrogen atom packing with two hydrogen atoms. We find if we derive the

equation that represents these computations, we have

Where,…

, ,

Radius of hydrogen atom

Remember our constant k equation 4.16 (Don’t forget to divide by two somewhere):

v =

0.833E − 15m

= 1.389E − 16m /s

= 1.2E − 10m

t =

1.2E − 10m

1.389E − 16m /s

= 863,930 . 8855s

= 431,965 . 4428s

= (773.5m /s)(431,965 . 4428s) = 334,125, 270m

334,125, 270m

299,792, 459m /s

= 1.11452193secon d s

v =

t = 6s =

h 4π r

= α

h 4π r

v =

= α

h 4π

= 1.2E − 10m

t =

= R

⋅

h 4π

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

of 15 70

Since we have the equation of the radius of a proton is given by, by evaluating it at one second

which is carbon:

And 1 second in terms of the atom is given by

Then the equation for the radius of a proton is:

Equation 1.22

And since

Then,…

Equation 1.23

Let’s verify equation 1.22

Remember that in equation 1.22, which is

3 2

⋅

Gπ

⋅

𝔼

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12secon d s

4πh

= 8.288587 × 10

−16

m = 0.829f m

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

2 ⋅

4π

⋅

𝔼

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

h 4π r

≈ 6 ⋅

K Eof Moon

K Eof Ear th

⋅ Ear th Da y = (6)1.2secon d s

3 2

⋅

Gπ

⋅

𝔼

K Eof Moon

K Eof Ear th

⋅ Ear th Da y

1.67E − 27

(6.626E − 34)(299,792, 459)

(6.674E − 11)4π

⋅

1.2E − 10

6.02E 23

= 0.93f m

of 16 70

That, must remain coupled. is determined by the number of protons in .

Making the approximation 9/8~1 we can write equation 1.22 as (We have picked up the fraction

9/8 by making several approximations)

Equation 1.26

Which gives

We form constants:

Equation 1.27

Equation 1.28

And we have the Equation:

Equation 1.29

We can say that Avogadro’s number is not an arbitrary number because it is such that there are

twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add

to this section the equation of the mass of a proton.

Equation 1.30

2 ⋅

4π

⋅

𝔼

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

(

)

(

)

(

)

ℍ

(

)

k =

2π

= 6.93E − 9kg

𝔼

= 1.99E − 34m

= k

𝔼

2π

⋅

𝔼

of 17 70

Discussion

Thus we have

Let us see if this makes sense.

is the circumference of a unit circle and is the area of a unit circle. For the constant of

gravitation

is a measure of the force of space pushing back on the proton per kilogram in our theory.

This is what G measures but over square meter per kilogram . Planck’s constant h is

The speed of light is m/s thus we have

Is a measure of force over surface or,…

Is a measure of energy over a length. Thus we see we have force canceling with force and area

with area to give us the radius of a proton:

We see we have force cancelling with force, surface area cancelling with surface area ,

leaving the square kilograms (kg) of the proton. We take the square root because we are forming

a geometric mean and then we divide by the mass of a proton and multiply by the radius of a

hydrogen atom to get the proper units for the radius of a proton. That is equation 1.26 makes

sense.

2π

⋅

𝔼

2π

= (2π)(π)(π)

2π

G =

kg ⋅ s

= N

⋅

h = kg

hc = kg

⋅

(

)

⋅ m

(

⋅ m

)

(

⋅

)

= kg

of 18 70

Section 2.0 The Charge of a Proton

of 19 70

Proton Charge

I construct the electric field as such (Fig. 2): is one component of . Events in

are through time with components and where c is the speed of light. The

electrons and the protons and in the presence of one another cross into giving

themselves acceleration in the and directions.

We have

We suggest

Equation 2.1

ℝ

(x, y, z)

ℝ

(x, y, z)

−x



0 ct

0 0 ct

= (ct

− ct

)



··

x ∝ (ct

− ct

)



Fig. 2

of 20 70

The position of changes in the space of and has travelled to a place in

time, as well. Since ct=meters there is a g such that . Thus since

and that is . Thus g is frequency squared ( ) and we suggest it is

derived from the separation between the charges and . That there is some velocity v such

that . We call upon our equation for 1 second:

Letting v= from our equation for k

And we have

Equation 2.2

We get

ℝ

(x, y, z)

t (0,ct

, ct

)

··

x = g(ct

− ct

)



g ⋅ m =

g =

⋅

⋅ s =

f =

(

)

h 4π

= 1secon d

(

)

(c)

h 4π

⋅

(c)

h 4π

= k

⋅

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= 773.5

m = m

(

⋅

)

h 4π r

= q

h 4π r

(6.626E − 34)(4π)(0.833E − 15)

(6.674E − 11)(299,792, 459)

= 5.37E − 31

q = (0.033)

36(18769

)

773.5

(5.37E − 31) = 9.1435E − 19C

of 21 70

Equation 2.3

Is carbon. We see the six-fold symmetry that is the carbon is due to the cube having 6

faces. I think I can be more accurate using space expanding in a sphere as opposed to a cube. We

have that

Equation 2.4

Equation 2.5

Equation 2.6.

Equation 2.7.

Equation 2.8.

Equation 2.9

Equation 2.10

Periodic table as cycles of six (3x6=18)

Hydrocarbons: The Skeltons of Life Chemistry

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6proton s

1E − 26kg ⋅ s

6secon ds

1E − 26kg ⋅ s

1secon d

18α

4πh

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6proton s

of 22 70

We said essentially at the outset if we consider it more explicitly that the periodic table is

structured the way it is because its cycle of 18 groups is founded in six-fold symmetry because 18

is 6 X 3=2 X 3 X 3 and 6 is 2 X 3 and 2 and 3 are the smallest prime numbers meaning no

number can be reduced further than factors of these, so these are the basis set at the crux of

Nature.

Ionization Energy

There was some idea I had that carbon and hydrogen are in inverse relationship to one another

in that we had!

Equation 1.14

We find six seconds gives 1 proton is hydrogen:

Equation 1.15

If we consider this something interesting happens like did with Giordano’s Relationship

because it had alpha and h combined to make G, just with the wrong units. But it turned out

that the Avogadro’s number solved that and turns out to be a natural unit, like we found the

second was. Now we have the same type of thing happening with the Joule, a unit of energy.

We find it useful to write out the constants in terms of their units we have used to describe

space, time, and matter and their properties in the previous sections…

, ,

We have that

Which is the units of the gravitational constant G is

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

G = N

= kg

⋅

kg ⋅ s

h = J ⋅ s = kg

⋅ m ⋅ s = kg ⋅

c =

= m

= kg

kg ⋅ s

(6.626E − 34)

1.6726E − 27E − 27)

⋅

299,792459

= 7.10048E 28N

of 23 70

Which is the units of force in Newtons is

The Bohr radius, or mean orbital radius of an electron in hydrogen for its ground state is

Is units of energy in Joules.

We want form a form of geometric mean with hydrogen (H) and carbon (C) in inverse

relationship to one another with respect time, t (the former multiplied by time the latter divided

by it. This associates the mass of a proton with one and the radius of a proton with the other; the

factor hc common to both:

Equation 2.11

The ionization energy of carbon is 1.804051E-18 Joules and that of hydrogen is 2.18E-18 Joules,

the energies to remove an electron (to move it from ground state to infinity). These are very

close to exactly the squares of our 1.4142E-9 Joules. Why would we have the square of the

ionization energies? Because the energy is not just over a line, but two lines that form an area,

and area is the square of a line. The ionization energy of hydrogen is given by the Rydberg

equation:

where and and A=2.18E-18J

The Electron affinity of carbon is 153.9kJ/mol

Electronegativity of carbon is 2.55

First ionization of carbon is 11.2603eV

IE=11.2603eV=1.8041E-18J.

We have (2.18E-18+1.8E-18)/2=1.99E-18 and

This is a form of the harmonic mean, Which is

= kg

(6.626E − 34)

(0.833E − 15m)

⋅

299,792459

= 286,274N

4πϵ

ℏ

= 5.29177E − 11m

= 7.1E − 5J

(

⋅ t

ℍ

)(

⋅

ℂ

)

= 1.4142E − 9kg

= 1.4142E − 9J = 2E − 9J

ΔE = A

(

−

)

= 2.18E − 18J

= ∞

= 1

C + IE → C

+ e

−

(1.414E − 9)

= 1.99E − 18

2(1.4142E − 9J )

(2.18E − 18J + 1.8E − 18J )

= 1Joule

of 24 70

Where A is approximately equal to B, and it is exactly one joule of energy which is the force of 1

newton acting through a meter.

2A B

A + B

= H

of 25 70

Section 3.0: The Planets And Atoms

of 26 70

Planets Mirror Atoms

We have our equation for the Earth-Moon-Sun Orbital System!

Equation 3.1.

And our equation for the proton

Equation 3.2.

Where is six protons is carbon (C). We also have that for the Earth orbital velocity

Equation 3.3.

This results in

Equation 3.4.

Which gives

Equation 3.5.

This is

Because k is in s/m is 1/k=773.5 m/s. Thus we have something like a quantum state on the left

but for the planets equal to a quantum state for the proton on the right, and conclude k connects

the larger scale macrocosmos to the microcosmos of the atom. Or perhaps better to write:

Equation 3.6.

Because then we have

Because . It makes one think of how the energy of a wave is the amplitude

squared. Here amplitude then is in seconds and energy is the square of time.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

α ⋅ 6m

h 4π r

= 1secon d

k v

≈ 6

K E

moon

K E

earth

⋅ Ear th Da y =

k v

h 4π r

(

K E

moon

K E

earth

Ear th Da y

)

⋅

h 4π r

k (m ⋅ s) = s

k v

(

K E

moon

K E

earth

Ear th Da y

)

⋅

h 4π r

= s

k v

⋅

= 1

of 27 70

Section 4.0 Proton-Seconds

of 28 70

Rigorous Formulation of Proton-Seconds

We can actually formulate this differently than we have. We had

But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are

lower and upper limits in an integral, then we have:

Equation 4.1

This Equation is the generalized equation we can use for solving problems. Essentially we can

rigorously formulate the notion of proton-seconds by considering

Equation 4.2

Is protons-seconds squared where current density is and ( can also be

). We say

Equation 4.3

Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a

number of protons. It is

Equation 4.4

Dividing Equation 7.2 through by t:

Equation 4.5

Which is proton-seconds. Dividing through by t again:

h 4π r

= 6proton s

h 4π r

= 1proton

h 4π r

∫

dt = ℕ

∫

qdt = t

∬

ρ(x, y, z)d x d y



J = ρ



ρ = Q /m

Q /m

Q =

∫

ρd V

ℕ =

h 4π r

∫

= t

∬

ρ(x, y, z)d x d y

of 29 70

Equation 4.6

We see that if where and then J is I/m2 (current per square meter)

is analogous to amperes per per square meter which are coulombs per second through a surface.

Thus we are looking at a number of protons per second through a surface. Thus we write:

Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to

protons in the output from our program and 1.0 seconds is carbon (C). We see we have the

following theorem:

Equation 4.7

So as an example,…

Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second

which is carbon and we have protons per square meter.

h 4π r

∫

= proton s



J = ρ



ρ = Q /m

v = m /s

h 4π r

∫

= 6

∫

1.0

0.5

−2

dt = − 6(1 − 2) = 6

h 4π r

∫

∬



J ⋅ d



h 4π r

∫

1.0

0.5

∬



J ⋅ d



S = − 3

(

1 −

0.25

)

= 9

proton s

secon d



J(x, y, z) = (0,0, J ) = − J



S = d x d y



J ⋅ d



S = (0,0, J ) ⋅ (0,0, d x d y) = − Jd x d y

of 30 70

We are now equipt to do computations in proton-seconds. We use equation 3.6 from two to

three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal

proton that we found described its radius (My feeling is we introduce the factor of 2 because

carbon is 6 protons +6 neutrons and 2 times 6 is twelve):

Now we integrate from phosphorus is 15 protons=0.396 seconds to aluminum is 13 protons =

0.462 seconds which is to integrate across silicon (divide your answer 10 by 2 to get protons):

By what value would you like to increment?: 0.006

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

251.2490 protons 0.024000 seconds 0.248978 decpart

59.1174 protons 0.102000 seconds 0.117416 decpart

50.2498 protons 0.120000 seconds 0.249805 decpart

40.1998 protons 0.150000 seconds 0.199844 decpart

37.2221 protons 0.162000 seconds 0.222080 decpart

27.1621 protons 0.222000 seconds 0.162058 decpart

25.1249 protons 0.240000 seconds 0.124907 decpart

20.0999 protons 0.300000 seconds 0.099916 decpart

17.0338 protons 0.354000 seconds 0.033823 decpart

16.2096 protons 0.372000 seconds 0.209604 decpart

15.2272 protons 0.396000 seconds 0.227202 decpart

14.1549 protons 0.426000 seconds 0.154862 decpart

13.2236 protons 0.456000 seconds 0.223620 decpart

13.0519 protons 0.462000 seconds 0.051885 decpart

12.1084 protons 0.498000 seconds 0.108375 decpart

11.1666 protons 0.540000 seconds 0.166615 decpart

11.0439 protons 0.546000 seconds 0.043906 decpart

10.1515 protons 0.594000 seconds 0.151471 decpart

Equation 4.8

Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus

enabling the construction of logic circuits that you can use to make computing machines. But

this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the

same group as aluminum, just above it. This results in a theory that there may be reason to

consider elements and their compounds as algebraic equation, not just chemical equations. We

show this now for silicon and germanium.

h 4π r

∫

= proton s

h 4π r

∫

= − 6(6.376 − 4.685) = 5protons /secon d = bor on

of 31 70

Theory of Compounds

Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the

elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and

germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain

noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four

electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope

the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two

types of silicon we have a semiconductor for making diodes and transistors from which we can

make logic circuits for AI.

As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the

right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.

I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy

in the differential because boron B is actually used to make positive type silicon.

That the AI elements make a three by three matrix they can be organized with the letter E with

subscripts that tell what element it is and it properties, I have done this:

Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the

first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be

organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in

three dimensional space so the mechanics of the realm we experience are described by such a

matrix, for example the cross product. Hence this paper where I show AI and biological life are

mathematical constructs and described in terms of one another.

We see, if we include the two biological elements in the matrix (E14) and and (E15) which are

carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to

show not only are the AI elements and biological elements mathematical constructs, they are

described in terms of one another. We see this because the first row is ( B, C, N) and these

happen to be the only elements that are not core AI elements in the matrix, except boron (B)

which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a

mathematical construct, the harmonic mean. Which means we have proved our case because the

first row if we take the cross product between the second and third rows are, its respective unit

vectors for the components meaning they describe them.

of 32 70

The Computation

And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological

elements C and N being the unit vectors are multiplied by the AI elements, meaning they

describe them. But we have to ask; Why does the first row have boron in it which is not a core

biological element, but is a core AI element? The answer is that boron is the one AI element that

is out of place, that is, aluminum is in its place. But we see this has a dynamic function.

The Dynamic Function

The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon

(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic

(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)

and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they

need doping agents from group 13 (three valence electrons) like boron and gallium, and to have

negative type silicon and germanium they need doping agents from group 15 like phosphorus

and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a

different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)

is in boron’s place. This results in an interesting equation.

Equation 4.9



A = (Al, Si, P)



B = (G a, G e, As)



A ×



B =



Al Si P

Ga Ge As

= (Si ⋅ As − P ⋅ G e)



B + (P ⋅ G a − Al ⋅ A s)



C + (Al ⋅ G e − Si ⋅ Ga)





A ×



B = − 145



B + 138



C + 1.3924



A = 26.98

+ 28.09

+ 30.97

= 50g /m ol

B = 69.72

+ 72.64

+ 74.92

= 126g /m ol



A ⋅



B = A Bcosθ

cosθ =

6241

6300

= 0.99

θ = 8

∘



A ×



B = A Bsin θ = (50)(126)sin8

∘

= 877.79

877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol

Si(As − G a) + G e(P − Al )

SiGe

Ge + Si

of 33 70

The differential across germanium crossed with silicon plus the differential across silicon

crossed with germanium normalized by the product between silicon and germanium is equal to

the boron divided by the average between the germanium and the silicon. The equation has

nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the

equation have a higher accuracy):

Due to an asymmetry in the periodic table of the elements due to boron we have the harmonic

mean between the semiconductor elements (by molar mass):

Equation 8.2

This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:

We can make this into two integrals:

Equation 4.10

Equation 4.11

28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)

(28.09)(72.61)

2(10.81)

(72.61 + 28.09)

0.213658912 = 0.21469712

0.213658912

0.21469712

= 0.995

(A s − Ga) +

(P − Al ) =

2SiG e

Si + Ge

0.213658912 = 0.21469712

0.213658912

0.21469712

= 0.995

∫

( ∇ × u ) ⋅ d S =

∮

u ⋅ d r

∫

[

(As − Ga) +

(P − Al )

]

d x dy ≈

Ge − Si

∫

xd x

∫

(As − Ga)dyd z ≈

(Ge − Si)

∫

xd x

∫

(P − Al )d x dz ≈

(Ge − Si)

∫

yd y

of 34 70

If in the equation (The accurate harmonic mean form):

Equation 4.12

We make the approximation

Equation 4.13

Then the Stokes form of the equation becomes

Equation 4,14

Thus we see for this approximation there are two integrals as well:

Equation 4.15

Equation 4.16

By making the approximation

We have

Equation 4.17

(As − Ga) +

(P − Al ) =

Ge − Si

2SiGe

Si + Ge

≈ Ge − Si

∫

[

(As − Ga) +

(P − Al )

]

dydz =

∫

d x

∫

(As − Ga)dyd z =

∫

(P − Al )dydz =

∫

2SiGe

Si + Ge

≈ Ge − Si

Si(As − Ga)

Ge(P − Al )

2SiGe

Si + Ge

ΔGe

ΔS

+ Ge

ΔSi

ΔS

= B

of 35 70

is the differential across Si, is the differential across Ge

and is the vertical differential.

We say (Phi) is given by

and

And

=1.618

=0.618

(phi) the golden ratio conjugate. We also find

Equation 4.18

We can go straight down group 14 to form the rest of our differentials:

It is amazing how accurately we can fit these differentials with an exponential equation

for the upward increase. The equation is

This is the halfwave:

Since B/Al=10.81/26.98=

0.40

And Ag/Cu =107.87/63.55=

ΔSi = P − Al

ΔGe = As − Ga

ΔS = Ge − Si

a = b + c

Φ = a /b

ϕ = b/a

(ϕ)ΔGe + (Φ)ΔSi = B

ΔC = N − B = 14.01 − 10.81 = 3.2

ΔSi = P − Al = 30.97 − 26.98 = 3.99

ΔGe = As − Ga = 74.92 − 69.72 = 5.2

ΔSn = Sb − In = 121.75 − 114.82 = 6.93

ΔPb = Bi − T l = 208.98 − 204.38 = 4.6

y(x) = e

0.4x

+ 1.7

y(x) = e

of 36 70

1.697

Where B and Al are the

differential across silicon

and Ag and Au are the finest

conductors of electricity, we

have

Equation 4.19

y(x) = e

(B/Al)x

of 37 70

6.0 The Planets As Six-fold Symmetry

of 38 70

Orbital Velocity

We can find our six-fold symmetry is characteristic of the solar system as well. The Earth rotates

through 15 degrees in one hour:

The distance then that the earth equatorial surface rotates through is where r is the earth

radius and theta is in radians.

Now we look at how many degrees through which the earth rotates in 1 minute:

For seconds…

Now let’s look at the same for the distance the earth moves around the Sun in an hour, a minute,

and a second as well…

And finally…

360

= 15

s = r θ

s = (6,378.14)(0.2618r a d ) = 1669.8k m /hr

24(60) = 1440min

360

1440

= 0.25

∘

= 0.0043633ra d

s = (6,378.14)(0.00436) = 27.83k m /min

24(60)(60) = 86,400s

360

∘

86,400

= 0.004167

∘

= 0.00007272ra d

s = (6,378,14)(0.00007272) = 0.464k m /s

(365.25)(24) = 8766hr /yr

360

8766

= 0.041

∘

= 0.00071678ra d

(149,598,000)(0.000716768) = 107,227k m /hr

(365.25)(24)(60) = 525960

360

525960

= = 0.000684463

∘

= 0.000011946ra d

s = (149,598,000)(0.000011946) = 1,787.1k m /min

of 39 70

We have the following table:

As we can see I am in good

agreement with Martin

Zombek, Handbook of Space

Astronomy and Astrophysics

which provides data. Notice

27.83 km/min in earth

rotation is approximately the

29.786 km/sec in earth orbit.

That is a valuable clue. Now let

us consider

Earth: 1 AU=149,598,000km (average earth-sun separation).

The earth rotates through about one degree a day.

Earth Orbit:

Earth Rotation: r=6,378.14km

Moon: sidereal month =27.83 days, r=405,400 km

(365.25)(24)(60)(60) = 31557600

360

31557600

= 0.000011408

∘

= 0.000000199ra d

(149,598,000)(0.000000199) = 29.786k m /sec

360

365.25

= 0.9856

deg

d a y

≈ 1

deg

d a y

360

∘

= 2π r a di a n s = 6.283

ra d

year

0.9856

∘

= 0.017202424

ra d

d a y

(149,598,000)(0.017202424) = 2,573,448.201

k m

d a y

360

∘

1d a y

= 6.283

ra d

d a y

s = r θ = (6,378.14)(6.283) = 40,075.0

k m

d a y

(27.83)(24) = 667.92

h ours

m onth

of 40 70

The points to be made in this exploration

1. So the earth goes through 1 degree a day and the moon 1 kilometer per second.

2. Earth rotates through 27.83 km/min at its equatorial surface and orbits through 29.786km/s

around the sun. These two are close to the same.

3. The synodic month is 29.53 days approximately equals the 29.786 km/s the Earth moves

around the sun. The sidereal month is 27.83 days is the 27.83 km/min through which the

Earth rotates at its surface.

With that information we have this mystery of sexagesimal in the earth-moon-sun orbital

parameters, solved. We see we can make the following equation in table form:

Consider minute times day:

Thus,..

The circle is divided into 360 units, each unit (each degree) is the distance the earth moves

around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are

360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).

We have:

1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always

close to that because its orbit is approximately circular. We have

This is approximately the diameter of the Earth orbit. We define our variables:

Earth orbits:

360

667.92

= 0.529

∘

= 0.0094

ra d

(405,400k m)(0.0094ra d /hr) = 3,810.76k m /hr

(3,810.75k m /hr )(hr /60min) = 63.5k m /min = 1.0585k m /s

orbit rotat ion orbit m oon

29.786k m minute 1da y kilom eter

secon d 27.83k m d egree secon d

min ⋅ d ay

deg

⋅

k m

(min)(d ay) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec

min ⋅ d ay

deg

⋅

k m

86,400s

deg

⋅

k m

= 86,400 ⋅

k m

deg

86,400 ⋅

k m

deg

⋅ 360

∘

= 311,040,000k m

311,040,000k m

149,598,000k m /AU

= 2.079AU

= 29.786

k m

of 41 70

Earth rotates:

Earth orbits:

Moon orbits:

Earth completes a 360 degree orbit yields:

Where on the right it is in radians and is the radius of the Earth’s orbit. We have

This is (0.00618)360=2.225

0.00618 is

Where is the inverse of the golden ratio. .

We have:

Equation 10.1

Our base ten counting is defined

is defined

, such that

which is given by

Thus since the diameter of the Earth orbit is

27.83k m

min

27.83k m

min

⋅

min

60sec

= 0.4638

k m

sec

1d a y

deg

(24)(60)(60)

deg

= 14,400

sec

deg

1k m

sec

⋅ θ

⋅ v

⋅

360

∘

1AU

= 2π

⋅ θ

⋅ v

⋅ r

= 1AU

(29.786)(14,400)(1k m /s)

(0.4638)(149,598,000)

360 = 2.225AU

100

ϕ = 0.618 =

100

360

∘

= 2AU

= 1,10

= 100,...

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

of 42 70

Then its radius is

Since we measure time with the Earth orbital period and that period is given by Kepler as

Then

Equation 10.2

Is approximately one year. In this section we set out two show the historical development of the

second by dividing up the motions of the Earth, Moon, and the apparent motion of the Sun into

units of 24, and 60 result in the solar system’s size as based around the Earth orbit giving us

Since we have established a connection between the microcosmos and the macrocosmos we

would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want

to know the universal gravitational constant in these units:

For a year we have

(365.25)(24)(60)(60)=31557600 seconds

100

360

∘

= 2AU

⋅

100

360

∘

= 1AU

= a

T =

10 ⋅

⋅ 360

∘

100

360

∘

= 2AU

G = 6.67408E − 11

kg ⋅ s

G = 6.67408E − 11

kg ⋅ s

⋅

3.3479E 33m

⋅

1.98847E 30kg

⊙

⋅

9.9477E14s

year

G = 39.433

⊙

⋅ year

G ≈ 40

⊙

⋅ year

⊙

= 40

year

⊙

4π

= 0.99885 ≈ 1

of 43 70

And for an AU

1 AU=1.496E11m

We can immediately put this to work. In order for the earth to stay in orbit its centripetal force

must equal the gravitational force. Its orbital velocity must be given by:

Since in our theory we present R as:

Then

Equation 10.3

Which evaluates:

Converting this to meters per second:

Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s

is an average over a varying velocity due to the Earth’s slightly elliptical orbit.

Since we have

For the earth, we show now it is true for the moon as well, and indeed it would hold for any

system circular, or approximately circular. For the Sun and its planets

If you use astronomical units and Earth years. In general we use

v =

⋅

100

360

∘

= 1AU

v =

(40)(1)

0.618

= 6

year

⋅

year

3.156E 7

⋅

1.496E11m

= 28,441

⋅

100

360

∘

= 1AU

= a

of 44 70

Which has the constant of proportionality

If m is small compared with M, you can write

Since we want to work with Earth and the Moon m is not so small compared to M. So,…

And these do round neatly to one-exponent like this. The sidereal lunar month is

(27.83)(24)(60)(60)=2,404,513 seconds. Thus,…

The tables give

Which is an average for the lunar orbit in its slightly elliptical orbit.

We need to find G in terms of Lunar Units (LU) as opposed to Astronomical Units (AU), and in

terms of the lunar month as one instead of the earth year. We see G has the same value we had

for solar-earth units:

= G

M + m

4π

= G

M + m

4π

= G

4π

= (6.674E − 11)

(

5.9722E 24kg + 7.347731E 22kg

4π

)

(6.674E − 11)

(

6E 24kg

4π

)

4E14

4π

1E14

= 1E13

= (1E13)(2,404,512)

= 5.78E 25

a = 3.8668E 8m

a = 3.833E 8m

G = 6.674E − 11

kg ⋅ s

⋅

(3.8668E8m)

⋅

5.9722E 24kg

⊕

⋅

(2,404,512s)

L M

40.896 ⋅

⊕

⋅ L M

of 45 70

We only need to convert to m, kg, s by using their relationships with LU, LM, and . That is,

for the orbital velocity is always six:

Where is the orbital velocity of the moon, and is the orbital velocity of the earth. The

difference comes when converting the system from lunar units in the case of the first to kg/m/s

and from astronomical units in the case of the second to kg/m/s. For example the first is:

Which is correct, the tables give that the lunar orbital velocity is 1000 m/s.

To find logos, is to find Nature not just as number, but as dimensionless whole

number, because fractions have inherent in them irrational numbers, which are

unending decimal expressions that at some point must be rounded to so many

figures to put them to use. Logos then is the language of Nature that transcends

conundrum because it would be utilized as relationship between point, plane, and

line as opposed to units. We have found that for the orbital velocity:

Equation 10.4

Where is the earth orbital radius. This results in the orbital velocity for a orbiting body is

Equation 10.5

Which is logos because it is number, which is 6. The orbital velocity of a body is always 6

because:

Where is the orbital velocity of the moon, and is the orbital velocity of the earth, That is G

is always 40, and M is always 1 providing the orbit is circular. Let us show this for Venus. Its

⊕

(40)(1)

= 6

(40)(1)

= 6

3.8668E8m

2,404,512s

= 964.886m /s

⋅

100

360

∘

= 1AU

v =

(40)(1)

= 6

(40)(1)

= 6

of 46 70

orbital distance VU (Venus units) is 1.082E11meters. It orbital period is 1.94E7s is the Venus

year (VY).

Thus we have

We can then express all orbital velocities as 6, but to find there values in a formal system of

units, we need to convert from these natural units to something like kg/m/s. Thus logos

translated into a language we understand can best be done in a square array, or as a matrix

transformation. We have

You will find this gives , ,

All of which are correct within the variations of these velocities in their deviations from a

perfectly circular orbit. Logos is:

Equations 10.6

G = 6.674E − 11

kg ⋅ s

⋅

(1.082E11m)

⋅

1.989E 30kg

⊙

⋅

(1.94E 7s)

V Y

39.44

V U

⊙

V Y

≈ 40

(40)(1)

= 6

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E 11m

31,557,600s

1.082E11m

19,400,000s

= 964.886m /s

= 28,443m /s

= 33,463.9m /s

(

)

⋅

100

360

∘

= 1LU

⋅

100

360

∘

= 1AU

⋅

100

360

∘

= 1V U

of 47 70

, such that

which is given by

To say that logos is:

Is to say that for the solar system Kepler’s Law of Planetary Motion holds:

That in general

Because as such for Earth we have

For the Moon we have

For Venus we have

v =

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E 11m

31,557,600s

1.082E11m

19,400,000s

= a

= G

M + m

4π

G = 6.67408E − 11

kg ⋅ s

⋅

3.3479E 33m

⋅

1.98847E 30kg

⊙

⋅

9.9477E14s

year

G ≈ 40

⊙

⋅ year

G = 6.674E − 11

kg ⋅ s

⋅

(3.8668E8m)

⋅

5.9722E 24kg

⊕

⋅

(2,404,512s)

L M

40.896 ⋅

⊕

⋅ L M

of 48 70

Making

, ,

This is derived from

Which is to say

Is logos because

such that

which is given by

G = 6.674E − 11

kg ⋅ s

⋅

V U

(1.082E11m)

⋅

1.989E 30kg

⊙

⋅

(1.94E 7s)

V Y

39.44

V U

⊙

V Y

≈ 40

(40)(1)

= 6

(40)(1)

= 6

(40)(1)

= 6

v =

⋅

100

360

∘

= 1AU

⋅

100

360

∘

= 1LU

⋅

100

360

∘

= 1AU

⋅

100

360

∘

= 1V U

v =

5 + 1

= Φ =

a = b + c

ϕ =

−

− 1 = 0

of 49 70

Which is the relationship between point, plane, and line.

From everything we have said G, the gravitational constant is about 40

We can do this for any planet and get G is approximately 40. We found the orbital velocity of any

planets is 6. This is true because as we have shown, the orbital distance of any planet is

This gives since

That orbital velocities are for the moon, earth, and venus

But what does this mean? It means since

And, v=6, that where the orbital velocities of the planets are 6, their distances from the sun are

all one.

This gives:

But where six is

G = 6.674E − 11

kg ⋅ s

⋅

(Ear th Dista nceMeters)

⋅

Sol arMa ssKilogra m s

⊙

⋅

(Ear thOrbit al Per io d Secon d s)

year

⋅

100

360

∘

= 1AU

v =

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E 11m

31,557,600s

1.082E11m

19,400,000s

R =

≈ 1

(

1 1 1

)

3.8668E8m

1.496E11m

1.082E11m

of 50 70

Since we have

And since we have said

Which we can write

Equation 10.7

Where r and T are orbital radii and orbital periods subscripted with their designations, then we

see Earth is the eigenvalue because

Equation 10.8

Which means

Or,…

Equation 10.9

That is since Eigenvalues are characteristic roots, that is the value such that for the matrix A:

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E 11m

31,557,600s

1.082E11m

19,400,000s

k v

= 6

(

6 6 6

)

3.8668E8m

2,404,512s

1.496E 11m

31,557,600s

1.082E11m

19,400,000s

(

6 6 6

)

(

6 k v

)

k v

= v



x = λ



of 51 70

Eigenvalues and Eigenvectors

The following matrix produces Eigenvalues 1 and 6…!

Since!

Implies!

Where!

Is the identity matrix. Then we see these are the the eigenvalues and , because!

Which is!

Which are the polynomials!

and !

Which means the eigenvectors are!

and !

Because!

and,…!

A =

(

4 1

6 3

)



u = λ



A − λ I =



I =

(

1 0

0 1

)

λ = 6

λ = 1

(

4 1

6 3

)

− λ

(

1 0

0 1

)

(

4 − λ 1

1 3 − λ

)



− 7λ + 6 = 0

(λ − 1)(λ − 6) = 0

y − 2x = 0

y + 3x = 0

(

)

(

−1

)

(

)

(

)

(

4 1

6 3

)(

−1

)

= 1

(

−1

)

of 52 70

The matrix!

Is a linear transformation for which the eigenvectors!

and !

Point in a direction in which the eigenvalue is stretched, and the eigenvalues!

and !

Are the factors by which it is stretched.!

(

4 1

6 3

)(

)

= 6

(

)

A =

(

4 1

6 3

)

(

)

(

−1

)

(

)

(

)

(

)

(

−1

)

(

)

(

)

of 53 70

Thus our transformation is!

Which is!

Thus everywhere in the planes below these

surfaces the vectors!

Point in the direction of the steepness of the

slopes of these planes respectively, and their

steepness are!

Respectively. We have done all of this so we

can visualize what is meant by the earth is the

eigenvalues for the matrix that is the matrix for

the orbits of the planets. In as much as we

have said:!

Which is to say



u ) =

(

4 1

6 3

)(

)

f (x, y) = 4x + y

g(x, y) = 6x + 3y

∇f (x, y) = 4



i + 1



∇g(x, y) = 6



i + 3



i + 1



i + 3



+ 1

= 17 = 4.123

+ 3

= 45 = 6.71

964.886m /s

28,443m /s

33,463.8m /s

of 54 70

Where we have taken the moon, Earth, and Venus as examples, but which works for any orbiting

body, and 6 is in!

And is !

We are essentially saying!

And we want to ﬁnd A. Since has no units, and on the right we must have!

Because these are the respective orbital velocities of the moon, Earth, and Venus, then!

That is!

And we have our eigenvectors are!

and. "



x = A



= A

964.886m /s

28,443m /s

33,463.8m /s

A =

160.814m /s

4,740.5m /s

5,577.3m /s

160.814m /s

4,740.5m /s

5,577.3m /s

(

)

(

)

of 55 70

We have!

And for the condition of eigenvectors to be satisﬁed by eigenvalues we have to have!

Which is to say!

And that is what we want to show. We have said that!

And, since !

And we showed that!

= A

λ = 6

A =

160.814m /s

4,740.5m /s

5,577.3m /s

= A

A =

160.814m /s

4,740.5m /s

5,577.3m /s

⋅

100

360

∘

= 1AU

v =

⋅

100

360

∘

= 1AU

of 56 70

We have shown that!

Which holds for all orbiting bodies if we use their natural units, like for Venus!

Where VU is Venus units as opposed to Earth Units (Astronomical Units). Thus for the Earth!

Because !

Meaning the units of phi are !

Since we have that!

Where means G in Natural System units means if your are working with earth you are

using astronomical units and earth years, if with Venus Venus units and so on. But so!

That the two eigenvectors!

v =

G = 6.67408E − 11

kg ⋅ s

⋅

3.3479E33m

⋅

1.98847E30kg

⊙

⋅

9.9477E14s

year

G ≈ 40

⊙

⋅ year

G = 6.674E − 11

kg ⋅ s

⋅

(1.082E11m)

⋅

1.989E30kg

⊙

⋅

(1.94E 7s)

V Y

40(1)

⊙

= 1

⊙

year

ϕ =

⊙

ϕ ≈ 5 /3

≈ G

⊙

= 1

= G

= 40 ≈ 6

of 57 70

Are equal and hence the condition for our eigenvectors!

Is satisﬁed because it can be written!

Equation 10.10. !

Where A is!

Equation 10.11. !

Which is what we want because we want!

Equation 10.12 !

Where:!

Which is

or Is Natural System Units!

= A

A =

160.814m /s

4,740.5m /s

5,577.3m /s

λ = k v

k =

8π

(1 + α) ⋅ N

𝔼

= 773.5m /s

(

)

= G

= 6

= 1

of 58 70

Conclusion

Indeed we have shown that we can predict the radius of a proton and its mass in terms

of the properties of space:!

Which gives

And the interesting thing is they do not yet have an equation for the radius of a proton, yet we

have it here and it doesn’t just use Planck’s constant, h, of the indeterministic microcosmos, but

uses the gravitational constant, G, of the deterministic macrocosmos. We have further shown

our theory of inertia is based on six-fold symmetry is centered around hydrocarbons, the

skeletons of life chemistry

We find 1 second gives

We find six seconds gives 1 proton is hydrogen:

We further find this same sixfold symmetry describes not just proton systems (atoms) but our

solar system

We also have that

And

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6 secon d s = hydrogen(H )

(

6 6 6

)

k v

= 6

(

⋅

)

h 4π r

= q

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6proton s

of 59 70

My feeling is if we want to go further to explain why six-fold symmetry we have to go into the

mathematics of symmetry breaking and perturbation theory. A particle has an equal problem of

falling either left or right out of an unstable equilibrium into wells on either to the left of it or

right. In the mathematics we say that even though there is an equal probability of going to the

left as opposed to the right or vice versa, one is magically chosen over the other, because we

don’t say a force exists that choses a direction, but use the word perturbation. But is a

perturbation a small force that has a preference as to which direction something will go?

Whether it it is, or not, is not clear because by quantum mechanics the particles does not have a

position, just a probability of existing in a given region at a given time. But does this mean if a

choice is made of falling in one direction or the other that there exists something physical going

on that we can explain that explains why the quantum mechanics is the way it is? Currently

quantum mechanics cannot describe anything deterministic, but only works in probabilities.

of 60 70

Appendix 1

We suggested the second was a natural unit, and that, if it was, should be in the orbital

mechanics of the earth because the second comes from the calendar, which is based on the

orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the

rotation of the Earth. We found that it was as the following:

Let’s show that…

To find the translational kinetic energy of the moon:

Distance from earth: 3.85E8m

Orbital period:

T=27.32 days=2.36E6seconds

v=1.025E3m/s~1000m/s

Mass: 7.34767E22kg

Use

E=3.67E28 Joules

To find the translational kinetic energy of the earth:

Distance from Sun: 1AU=1.496E11m

Orbital period: 1 year=3.1558E7 seconds

v=2.9785E4m/s, Earth mass: 5.9722E24kg

E=2.649E33 Joules

Earth day=(24)(60)(60)=86,400 seconds.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

K . E . Moon = 3.67E 28J

K . E . Ear th = 2.649E 33j

2π (3.85E8m) = 2.419E 9m

E =

2π (1.496E11m) = 9.399E11m

of 61 70

Appendix 2: Sexagesimal And The Cosmic Calendar

My book Abstract Cosmology could have easily been called The Book of Six, it was

Buckminster Fuller who said “Nature employs 60 degree coordination”. Sixty degrees are the

degrees of an equilateral triangle. The triangle is the structure that encloses an area with the

least amount of sides. Buckminster Fuller said, systems of triangles are the the only inherently

stable patterns. The regular hexagon, a six-sides polygon tessellates as equilateral, 60 degree,

inherently stable equilateral triangles. It was the scientist Shubnikov who said among the living

organisms the pattern with which we most frequently meet is ﬁve-fold symmetry. It is well

known that the physical, like snowﬂakes, are six-fold symmetry. Thus we should not be

surprised that our Abstract Cosmology is founded on six-fold symmetry. Two and three are the

smallest prime numbers and their product is six. Two times six is twelve, the number most

evenly divisible by whole numbers for its size (it is a so-called abundant number: divisible

evenly by 1,2,3,4,6, their sum is 16 which is greater than twelve itself and, three times six is

eighteen, the cyclical Nature of the periodic table (18 groups). !

At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and

hunting, to invent agriculture and and build ceramic homes, developed the ﬁrst mathematics,

and it was this sexagesimal (base 60) that passed on to the Babylonians, then ended up with

the Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into

60 seconds of time, but who divided the sky into hours, minutes, and seconds of arc which

was measured in time counted by the rotation of the Earth, and its orbital period around the

Sun, and the orbital period of the Moon around the Earth.!

We are suggesting nature is founded on six-fold symmetry, six which is constructed by

multiplying together the ﬁrst two prime numbers, 2 and 3. Two factorial is two, three factorial is

6. Two times six is twelve, the number of months in a year, approximated by our moon’s

approximately 12 orbits around the earth in the time it takes the earth to go around the sun

once. There are four weeks in a lunar month, and!

Is the number of seconds in the 24 hour Earth day. The duration of one second comes from

dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,

and Ancient Greeks and it is believed they did this because 60 is evenly divisible by!

1,2,3,4,5,6,…12,15,20,30,60,…!

The moon orbits the Earth in approximately 30 rotations of the Earth, hence the 30 day month.

January has 31 days, February 29, March 31, April 30 May 31, June 30, July 31, August 31,

September 30, October 31, November 30, December 31. This averages out to!

We divide a circle into 360 units called degrees, which is six squared times ten. As such the

equilateral triangle has each angle equal to 60 degrees. There are approximately 360 earth

rotations in the time it orbits the sun once, Hence in one day the Earth moves through

approximately one degree in one day in its 365 day journey around the Sun. The sidereal

month, the time it takes the moon to return to the same position against the background of the

stars is 27 days 7 hours 43 minutes. We see the power of sexagesimal (base 60) in computing

time with the month divided into days, hours, and seconds where there are 60 minutes in an

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60

= 86,400

31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + 31

367

= 30.58333

of 62 70

hour and 60 seconds in a minute because we can convert this easily into seconds with such a

system as here, the duration of this sidereal month in seconds:!

The length of the synodic month, the completion of the phases of the moon, or the time it takes

the moon to return to the same position with respect to the Sun, is 29 days 12 hours 44

minutes 3 seconds. It is a little longer than the sidereal month because the earth has moved

with respect to the sun by the time the moon has come back around to a full orbit. We have:!

We looked at the arithmetic mean of the days in the month over a year, Let us look at the

harmonic mean:!

Gathering like terms in the denominator:!

=0.225806+0.034482759+0.133333=0.39362!

Which is the most frequent value for the days in a month. Now let’s look at the geometric mean

which tempers the various values with one another.!

=30.493!

But perhaps it is more telling to take the mean between the individual days of the month that

we use:!

0.032258+0.034482759+0.033333=0.10007!

(24 ⋅ 27 + 7) ⋅ 60

+ 43 ⋅ 60

+ 0 ⋅ 60

= 2,358,000 + 2,580 + 0 = 2360580sec /m onth

(24 ⋅ 29 + 12) ⋅ 60

+ 44 ⋅ 60

+ 3 ⋅ 60

= 2,548,800 + 2,640 + 3 = 2551443sec /m onth

0.032258 + 0.034482759 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258 + 0.032258 + 0.033333 + 0.032258 + 0.033333 + 0.032258

0.39362

= 30.486

31 ⋅ 29 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 31 ⋅ 30 ⋅ 31 ⋅ 30 ⋅ 31

31 + 29 + 30

= 30

0.10007

= 29.799 ≈ 30

31 ⋅ 29 ⋅ 30 = 29.98888 ≈ 30

of 63 70

We see that the arithmetic mean gives us exactly a 30 day month, hence we always speak

loosely saying a month is 30 days. !

Indeed when Buckminster Fuller speaks of Nature employing 60 degree coordination he is

speaking of the equilateral triangle as the fundamental unit of one for area. That is the six-sided

regular hexagon can be divided into six equilateral triangles, giving it an area of six. Indeed he

takes the equilateral triangle as one because not only does it enclose an area with the fewest

sides, it has the smallest area for its perimeter; a regular hexagon with a perimeter the same

size encloses more area. I would like to say the Ancient Sumerians, Babylonians, and Greeks

made this connection, because in their sexagesimal (base 60) system of notation, one is a

straight line with a triangle on top. See ﬁgure below. Ancient Mesopotamian Cuneiform for

writing. It is not sexagesimal notation in the strict sense, as they have a unique symbol for a

tens column, but it was this sexagesimal structure that lead to the 60 minute hour and 60

second minute, as well as the angular division projected onto the sky in 60 minutes of arc in a

degree and 60 seconds of arc in a minute as a way of mathematically deﬁning time dynamically

in terms of the motions of the Earth and moon.!

For equation our equation Earth day needs to be shorter. A long time ago it

was; the Earth loses energy to the moon. The days become longer by 0.0067

hours per million years. Our Equation

Is actually 1.2 seconds.

K E

moon

K E

earth

(Ear th Da y) ≈ 1secon d

of 64 70

We have

24-20=0.0067t

t=597 million years

This was when the earth went through a dramatic change and there was a big explosion of life

(The Cambrian). The dinosaurs went extinct 65 million years ago giving small mammals a

chance to evolve paving the way for humans.

24-x=0.0067t

x=23.5645 hours

We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic

calendar:

What is the next term?

20+3+0.57735+0.4714=24hours

Which bring us to today.

It would seem we could have divided that day, not into 24 hours but 20 hours and get our 1.2

seconds as a duration for measuring time:

Both work because

24=2x2x6

20=>20+20+20=60

Both have an intimate structural relationships with sexagesimal.

24h ours

1.2

= 20h ours

3cos(0

∘

) +

cos(30

∘

) = dinosa ur − ex t inct ion =

20hrs + 3hrs +

hrs +

(365.25d a ys)(24h ours)(60sec)(60sec)

(365.25d a ys)(20h ours)(60sec)(60sec)

= 1.2

of 65 70

Appendix 3

Let us start with the units with which we are working:

And convert these to proton-masses and proton-radii:

Now we find k in these units:

Thus we have:

Thus we see times is approximately one mole, that is the radius of one hydrogen atom

times k is approximately one mole.

Equation 6.1.

So we have

G =

kg ⋅ s

h = kg ⋅

c = m /s

G = 6.67408E − 11

kg ⋅ s

⋅ 1.67262E − 27kg(0.833E − 15m)

⋅ s = 193,131,756

h = 6.62607E − 34kg ⋅

⋅

(0.833E − 15)

(1.67262E − 27kg)

= 5.71E 23

c =

(299,792,459m /s)(1sec)

(0.833E − 15m)

= 3.6E 23

1.2E − 10m

0.833E − 15m

= 144,058

k =

2π

= 6.93E − 9kg

k =

(5.71E 23)(3.6E 23)

2π

(193131756)

= 4E18proton − m a sses

= k

𝔼

(4E18)(144058)

(6E 23)

5.76E 23

(6E 23)

= 0.96 ≈ 1

k = 1m ole

of 66 70

We can say the molar mass of hydrogen is

This is really what we are saying when we are speaking of the molar mass of hydrogen, that there

is one atom of hydrogen per proton by mass. What we have discovered is

Which is to say mass is inverse relationship to space, i.e. a length is the inverse of a mass.

This by a factor of

Indeed the radius of a proton to the mass of a proton needs to be exactly 1.00 not 0.96, but the

0.96 is close enough to work with. However, the radius of a hydrogen atom is vague and the

value of 1.2E-10 is a construct that satisfies certain models. We can suggest that 144058 is

actually 150,000, in which case (150,000)(4E18) is exactly 6E23 or Avogadro’s number. If we do

this then it is to say the radius of a hydrogen atom is 1.2495E-10m. This could be some sort of a

functional radius in terms of the mechanics of what is going on here, it does not mean 1.2E-10 is

not a functional radius in terms of other mechanical systems. That which we want to do is look

at equation 6.4 so we can understand if the radius of a proton as the inverse of its mass, is

compatible with our model of inertia in fig. 1 which determines equations equations 4.3 and 4.7.

We see then, that a hydrogen atom is 150,000 times bigger than its proton:

2π

= 6E 23m

6E 23 ⋅

atom sH

gr a m

⋅ 1,67262E − 24

gr a m s

proton ⋅ m a ss

1.00

atom sH

proton ⋅ m a ss

⋅

(1.2E − 10m)(6.93E − 9kg)

(6E 23)

= 1.386E − 42

(144058r

)(4E18m

)

(6E 23))

= 0.96

= 1

2π

⋅

𝔼

2π

⋅

𝔼

1.2495E − 10 m

0.833E − 15m

= 150,000

of 67 70

And here is the innovation: We usually think of the building blocks forming first, like the

protons, as building blocks for something of which they are a part, like hydrogen atoms, which

are a proton with an electron orbiting them. But I do not think reverse thinking should be over-

looked, that it may be reasonable to suggest that protons are not necessarily the cause of atoms,

like the hydrogen atom here, but that atoms are the cause of protons, that protons formed to

accommodate atoms, not vice versa. In which case we want to now look at the volume of a

hydrogen atom, and find how many protons and neutrons fit in it if its density is to equal

carbon. We look at carbon because we are concerned with our equations of hydrocarbons, the

skeletons of life that are characteristic of our inertia equations

We find six seconds gives 1 proton is hydrogen:

Firstly, our hydrogen volume equation is:

The density of carbon is:

We write

With the mass of a proton equal to the mass of a neutron this is carbon-11 also called carbane

which is unstable and decays with a half-life of 20.4 minutes to Boron-11. Boron 11 is not only

stable, but is boron’s most abundant isotope at 80.1% of the atoms in any natural sample. The

other stable isotope is Boron-10 at 19.8%. Boron is just before carbon in the periodic table of

elements giving it atomic number 5. Thus carbon-12, around which molar mass was defined for

all of the elements is 6 protons and 6 neutrons. My suggestion here that in order for a

hydrocarbon, such as the simplest, methane CH4, must require a hydrogen atom to have a

volume on the order of fitting carbons 6 protons and 6 neutrons. We see here it does, and that it

does especially in light of a hydrogen atoms volume is ambiguous in that its size is described in

terms of the distance of its electron from its proton, which changes as the electron jumps from

orbit to orbit. In fact carbon-11 does fit exactly in this definition for a radius for hydrogen. But it

decays into boron-11 which does then fit perfectly, and we have seen that boron plays a pivotal

roll in our theory of algebraic elements.

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6 secon d s = hydrogen(H )

π R

π (1.2495E − 10)

= 8.1714173E − 30

2.26

= 2,260

2,260 = n

1.67262E − 27kg

8.17E − 30m

n = 11protons

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Appendix 4: The Constant k Neutron Star

We see our constant k works with the gravitational constant G to predict the radius of a

proton, and with the electric constant to predict the charge of electron as a result of

the six-fold symmetry of the core life element carbon. It follows from our idea of an

intermediary mass!

Where M is given by the Chandrasekhar limit for a white-dwarf star not to collapse:

Giving for k

Which is 1/k=773.5 m/s. There is another kind of star we can use as the limiting factor

for determining a constant k with a different value, a so-called neutron star. The

Chandrasekhar limit is determined by the idea that it resists collapse from electron

degeneracy pressure which is the only thing countering its gravity because it does not do

nuclear fusion. By the Pauli exclusion principle there is pressure from electron

degeneracy because it says that two or more identical particles with half integer spins

cannot occupy the same quantum state within a quantum system simultaneously, which

means the electrons are in motion.

A neutron star is the smallest and densest star we know of because it is made of

neutrons right up against one another giving them near the density of a neutron. They

are the collapsed core of a supergiant from masses of ten to twenty-five solar masses.

They resist collapse by neutron degeneracy pressure in contrast to white dwarves with

electron degeneracy pressure by the same Pauli exclusion principle. While we have an

equation predicting the upper-limit for their mass, it is only a ballpark figure because we

don’t know in quantum mechanics what the properties of neutrons are at such

pressures, however we have been able to experimentally get a value that is thought to be

accurate by observations of actual neutron stars. That value is (About two solar

masses). If its mass exceeds that the neutron star will collapse into a blackhole. Thus we

have another value to obtain an intermediary value for mass, and hence another k. We

will call it . We have

= Mm

M ≤ 0.77

ℏ

= 1.41 ⊙

k =

8π

(1 + α) ⋅ N

𝔼

2.16 ⊙

= 1.67262E − 27)(1.989E 30)(2.16) = 84.770kg

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This in contrast to for a white dwarf of 69.205 kg. That is a ratio of 1.2. We have the

value of

In contrast to

The great thing about working with the white dwarf star was that we have an equation

for it, the Chandrasekhar limit so we have a more complete equation for k in that

scenario

That is we are in that scenario not limited to

But the result of k for a neutron star does have some interesting integer properties with the

planets. Where k produced close to the six of our sixfold symmetry using the orbital velocity of

the earth

We have even closer integers with many of the planets for . But we don’t need to go into that

here.

(6.62607E − 34)(1.0073)

6.67409E − 11

(6.02E 23)

84.770

= 0.0083779s /m

= 11.936

≈ 12

= 773.5

k =

8π

(1 + α) ⋅ N

𝔼

k =

(1 + α)

⋅ N

𝔼

k v

= 6

of 70 70

The Author